Explain why ${\left[Fe{\left(H}_{}\right]}^{}$20)63+ has magnetic moment value of 5.92 BM whereas

${\left[Fe{\left(CN\right)}_{6}\right]}^{3-}$ has a value of only 1.74 BM?

Ans/.

${\left[Fe{\left(CN\right)}_{6}\right]}^{3-}$involves ${d}^{2}s{p}^{3}$hybridisation with one unpaired electron (as shown by its

magnetic moment 1.74 BM) and ${\left[Fe{\left({H}_{2}0\right)}_{6}\right]}^{3+}$involves sp${}^{3}{d}^{2}$hybridisation with five

unpaired electrons (because magnetic moment equal to 6.92 BM)

CN${}^{-}$ is stronger ligand than H${}_{2}$,O according to specttochemical series.${∆}_{0}>P$ for CN${}^{-}$hence,

fourth electron will pair itself, Whereas for water pairing will not happen for ${\left[Fe{\left(CN\right)}_{6}\right]}^{3-}$ the

electronic configuration of Fe${}^{3+}$ is

One unpaired electron

For ${\left[Fe{\left(H}_{}\right]}^{}$2O)63+ the electronic configuration of Fe${}^{3+}$ is

Five unpaired electron

Hence, ${\left[Fe{\left(CN\right)}_{6}\right]}^{3-}$ and ${\left[Fe{\left({H}_{2}O\right)}_{6}\right]}^{3+}$ are inner obital an douter orbital complex respectively